April 3 Math 2335 sec 001 Spring 2014

Section 5.1: Trapezoid and Simpson’s Rules

I(f ) =∫ b

af (x)dx

Trapezoid Rule: If [a,b] is divided into n equally spaced subintervalsof length h = (b − a)/n, then I(f ) ≈ Tn(f ) where

Tn(f ) =h2[f (x0) + 2f (x1) + 2f (x2) + · · ·+ 2f (xn−1) + f (xn)]

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Trapezoid and Simpson’s Rule

I(f ) =∫ b

af (x)dx

Simpson’s Rule: If [a,b] is divided into an even number n equallyspaced subintervals of length h = (b − a)/n, then I(f ) ≈ Sn(f ) where

Sn(f ) =h3[f (x0) + 4f (x1) + 2f (x2) + 4f (x3) + 2f (x4)+

+ · · ·+ 2f (xn−2) + 4f (xn−1) + f (xn)].

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Example: Approximate I(f ) with T4, S2 and S4Compute the error |I(f )− Xn(f )| for each case. (X = T ,S and n = 2,4)

I(f ) =∫ π

2

0

dx1 + sin x

= 1

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Section 5.2: Error in Tn and Sn1

Consider the partiaion a = x0 < x1 < · · · < xn = b of equally spacednodes with h = xj+1 − xj . On the subinterval [xj , xj+1] we have the errorformula for P1 (assuming f ′′ exists)

f (x)− P1(x) =(x − xj)(x − xj+1)

2f ′′(cj) for some xj ≤ cj ≤ xj+1.

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Error Formula for TnCompute the integral.∫ xj+1

xj

(x − xj)(x − xj+1)2

dx =12

∫ h0

u(u − h)du where u = x − xj

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Error ETn Formula for TnWriting

h3

12=

(h2

12

)h

we can add the errors from [x0, x1], [x1, x2], etc. together to get

ETn (f ) = −h2

12[hf ′′(c0) + hf ′′(c1) + · · ·+ hf ′′(cn−1)

]

We’re using E for error with a superscript T for the rule and subscript nfor the number of subintervals.

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Error ETn Formula for Tn

Recognizing the expression in brackets [ ] as a Riemann sum, we have

ETn (f ) = −h2

12[hf ′′(c0) + hf ′′(c1) + · · ·+ hf ′′(cn−1)

]

≈ −h2

12

∫ ba

f ′′(x)dx = −h2

12(f ′(b)− f ′(a)

)

By the Mean Value Theorem

f ′(b)− f ′(a) = (b − a)f ′′(c), for some a ≤ c ≤ b.

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Error ETn Formula for Tn

Our final error formula is

ETn (f ) = I(f )− Tn(f ) = −h2

12(b − a)f ′′(c)

for some c between a and b.

Since h = (b − a)/n, we can express ETn in terms of n as

ETn (f ) = −(b − a)3

12n2f ′′(c).

We see that the error is proportional to 1n2 .

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ExampleEstimate the number of subintervals needed to guarantee an accuracy

|ETn (f )| ≤ 10−4

for the integral

I(f ) =∫ 1

0

dx1 + x

.

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Error in Trapezoid Rule

I(f ) =∫ 1

0

dxx + 1

= ln 2

n Tn ETn2 0.70833333 −0.015186154 0.69702381 −0.003876638 0.69412185 −0.0009746716 0.693391202 −0.00024402

Since ETn ∝ 1n2 , doubling the number of subintervals reduces the errorby a factor of 4.

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Error in Simpson’s Rule ESn

A similar derivation shows that for f sufficiently differentiable,

ESn (f ) = I(f )− Sn(f ) = −h4

180(b − a)f (4)(c)

for some c between a and b. Since h = (b − a)/n, the above can bewritten as

ESn (f ) = −(b − a)5

180n4f (4)(c)

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ExampleEstimate the number of subintervals needed to guarantee an accuracy

|ESn (f )| ≤ 10−4

for the integral

I(f ) =∫ 1

0

dx1 + x

.

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Error in Simpson’s Rule

I(f ) =∫ 1

0

dxx + 1

= ln 2

n Sn ESn2 0.69444444 −0.001297264 0.69325397 −0.000106798 0.69315453 −0.00000735

16 0.69314765 −0.00000047

Since ESn ∝ 1n4 , doubling the number of subintervals reduces the errorby about a factor of 16.

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Order of a Rule

Note that the errors have the form

Error ≈ cnp

for some numbers p and c. In particular

ETn (f ) =cn2, and ESn (f ) =

cn4

for many functions f .

We may refer to p as the order of the integration rule.

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Richardson ExtrapolationLet an integration rule be denoted by In and have order p. TheRichardson extrapolation formula for this integration rule is

R2n =1

2p − 1(2pI2n − In

).

In particular, for the Trapezoid rule

R2n =13(4T2n − Tn) .

In particular, for Simpson’s rule

R2n =1

15(16S2n − Sn) .

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ExampleUse the previous results for the Trapezoid rule to compute R4(f ) for∫ 1

0

dxx + 1

and compare the error of R4 to that of T8.

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